I don’t understand this at all and I hate stats plz help

Image shown down below

faust18 [17]

Answer:

there is no answer down below

7 0

4 years ago

What value goes in box A?

marishachu [46]

Step-by-step explanation:

Hello!

For this you multiply the numbers that intersect at that box

Box A 2x and 4 intersect

2x * 4 = 8x

Box B 3 and 4 intersect

3 * 4 = 12

Box C 2x and x intersect

2x * x = $2x^{2}$

Box D 3 and x intersect

3 * x = 3x

To solve this we put all these together

$2x^{2} +3x+8x+12$

Combine like terms

$2x^{2} +11x+12$

Hope this helps!

5 0

3 years ago

Tyrell says $750 is about $800. Sara says $750 is about $700. who is correct? Explain

Dima020 [189]

Tyrell is correct.. If you round it, it is 750. 5 or more up the score, 4 or less let it rest.

6 0

3 years ago

The following table shows scores obtained in an examination by B.Ed JHS Specialism students. Use the information to answer the q

Makovka662 [10]

Answer:

(a) The cumulative frequency curve for the data is attached below.

(b) (i) The inter-quartile range is 10.08.

(b) (ii) The 70th percentile class scores is 0.

(b) (iii) the probability that a student scored at most 50 on the examination is 0.89.

Step-by-step explanation:

(a)

To make a cumulative frequency curve for the data first convert the class interval into continuous.

The cumulative frequencies are computed by summing the previous frequencies.

The cumulative frequency curve for the data is attached below.

(b)

(i)

The inter-quartile range is the difference between the third and the first quartile.

Compute the values of Q₁ and Q₃ as follows:

Q₁ is at the position:

$\frac{\sum f}{4}=\frac{100}{4}=25$

The class interval is: 34.5 - 39.5.

The formula of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 25 = 34.5

(CF) $_{p}$ = cumulative frequency of the previous class = 24

f = frequency of the class interval = 20

h = width = 39.5 - 34.5 = 5

Then the value of first quartile is:

$Q_{1}=l+[\frac{(\sum f/4)-(CF)_{p}}{f}]\times h$

$=34.5+[\frac{25-24}{20}]\times5\\\\=34.5+0.25\\=34.75$

The value of first quartile is 34.75.

Q₃ is at the position:

$\frac{3\sum f}{4}=\frac{3\times100}{4}=75$

The class interval is: 44.5 - 49.5.

The formula of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

Here,

l = lower limit of the class consisting value 75 = 44.5

(CF) $_{p}$ = cumulative frequency of the previous class = 74

f = frequency of the class interval = 15

h = width = 49.5 - 44.5 = 5

Then the value of third quartile is:

$Q_{3}=l+[\frac{(3\sum f/4)-(CF)_{p}}{f}]\times h$

$=44.5+[\frac{75-74}{15}]\times5\\\\=44.5+0.33\\=44.83$

The value of third quartile is 44.83.

Then the inter-quartile range is:

$IQR = Q_{3}-Q_{1}$

$=44.83-34.75\\=10.08$

Thus, the inter-quartile range is 10.08.

(ii)

The maximum upper limit of the class intervals is 69.5.

That is the maximum percentile class score is 69.5th percentile.

So, the 70th percentile class scores is 0.

(iii)

Compute the probability that a student scored at most 50 on the examination as follows:

$P(\text{Score At most 50})=\frac{\text{Favorable number of cases}}{\text{Total number of cases}}$

$=\frac{10+4+10+20+30+15}{100}\\\\=\frac{89}{100}\\\\=0.89$

Thus, the probability that a student scored at most 50 on the examination is 0.89.

5 0

4 years ago

How can you solve 25% of 44 both mentally and with pencil and paper?

STALIN [3.7K]

Answer:

<u>Solving it mentally</u> :-

It says 25% × 44. When we have a percent, change it to decimal and solve which would be easy to do it mentally.

25% = 0.25

0.25 × 44 = 11 *Answer is bolded.*

<u>Solving it on paper</u> :-

I have attached a picture of me doing it on paper.

Hope this helps, thank you :) !!

3 0

3 years ago